Introduction

In this blog post, we will explore how to solve the LeetCode problem "955" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given an array of n strings strs, all of the same length. We may choose any deletion indices, and we delete all the characters in those indices for each string. For example, if we have strs = ["abcdef","uvwxyz"] and deletion indices {0, 2, 3}, then the final array after deletions is ["bef", "vyz"]. Suppose we chose a set of deletion indices answer such that after deletions, the final array has its elements in lexicographic order (i.e., strs[0] <= strs[1] <= strs[2] <= ... <= strs[n - 1]). Return the minimum possible value of answer.length. Example 1: Input: strs = ["ca","bb","ac"] Output: 1 Explanation: After deleting the first column, strs = ["a", "b", "c"]. Now strs is in lexicographic order (ie. strs[0] <= strs[1] <= strs[2]). We require at least 1 deletion since initially strs was not in lexicographic order, so the answer is 1. Example 2: Input: strs = ["xc","yb","za"] Output: 0 Explanation: strs is already in lexicographic order, so we do not need to delete anything. Note that the rows of strs are not necessarily in lexicographic order: i.e., it is NOT necessarily true that (strs[0][0] <= strs[0][1] <= ...) Example 3: Input: strs = ["zyx","wvu","tsr"] Output: 3 Explanation: We have to delete every column. Constraints: n == strs.length 1 <= n <= 100 1 <= strs[i].length <= 100 strs[i] consists of lowercase English letters.

Explanation

Here's the breakdown of the problem and the solution:

• Approach:

  • Iterate through all columns of the input strings.
  • For each column, check if deleting it would make the strings lexicographically sorted up to that column. If not, increment the deletion count.
  • Maintain a boolean array to track whether the strings are sorted up to a given string.

• Complexity:

  • Runtime: O(n * m), where n is the number of strings and m is the length of each string.
  • Space: O(n), for the sorted_upto array.

Code

    def minDeletionSize(strs: list[str]) -> int:
    """
    Calculates the minimum number of columns to delete to make the strings lexicographically sorted.
    """
    n = len(strs)
    m = len(strs[0])
    ans = 0
    sorted_upto = [True] * n

    for j in range(m):
        temp_sorted_upto = sorted_upto[:]
        for i in range(1, n):
            if sorted_upto[i]:
                if strs[i - 1][j] > strs[i][j]:
                    ans += 1
                    break  # Move to the next column

        else:  # If no break occurred, the column is "good"
            for i in range(1, n):
                if sorted_upto[i] and strs[i - 1][j] < strs[i][j]:
                    continue
                elif sorted_upto[i] and strs[i - 1][j] == strs[i][j]:
                      continue
                else:
                    temp_sorted_upto[i] = False  # String no longer sorted

            sorted_upto = temp_sorted_upto[:]

    return ans